Let A be `3xx3` symmetric invertible matrix with real positive elements. Then the number of zero elements in `A^(-1)` are less than or equal to :

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the number of zero elements in the inverse of a \(3 \times 3\) symmetric invertible matrix \(A\) with real positive elements.

### Step-by-Step Solution:

1. **Understanding the Matrix Properties**:
- A \(3 \times 3\) symmetric matrix has the form:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{pmatrix} \]
- Since \(A\) is invertible and has real positive elements, all its eigenvalues are positive.

2. **Using the Formula for Zero Elements**:
- The number of zero elements in the inverse of a matrix can be bounded by the formula:
\[ \text{Number of zero elements in } A^{-1} \leq x^2 - 2x \]
- Here, \(x\) is the degree of the matrix, which in this case is \(3\).

3. **Calculating the Bound**:
- Substituting \(x = 3\) into the formula:
\[ \text{Number of zero elements in } A^{-1} \leq 3^2 - 2 \cdot 3 \]
- Calculating this gives:
\[ 3^2 = 9 \quad \text{and} \quad 2 \cdot 3 = 6 \]
- Therefore:
\[ 9 - 6 = 3 \]

4. **Conclusion**:
- The maximum number of zero elements in \(A^{-1}\) is \(3\). Thus, we conclude that the number of zero elements in \(A^{-1}\) is less than or equal to \(3\).

### Final Answer:
The number of zero elements in \(A^{-1}\) is less than or equal to \(3\).

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